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4 months ago

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You drag a crate across a floor with a rope. The force applied is 750 N and the angle of the rope is 25.0° above the horizontalH

ow much work do you do on the crate when moving it 26.0 m?If you complete the job in 6.00 s, what power is developed?

1 answer:

frez [133]4 months ago

5 0

a.

The work done by a constant force along a rectilinear motion when the force and the displacement vector are not colinear is given by:

$W=F\cos\theta\cdot d$

where F is the magnitude of the force, theta is the angle between them and d is the distance.

The problen gives the following data:

The magnitude of the force 750 N.

The angle between the force and the displacement which is 25°

The distance, 26 m.

Plugging this in the formula we have:

$\begin{gathered} W=\left(750\right)\left(\cos25\right)\left(26\right) \\ W=17673 \end{gathered}$

Therefore the work done is 17673 J.

b)

The power is given by:

$P=\frac{W}{t}$

the problem states that the time it takes is 6 s. Then:

$\begin{gathered} P=\frac{17673}{6} \\ P=2945.5 \end{gathered}$

Therefore the power is 2945.5 W

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The membrane that surrounds a certain type of living cell has a surface area of 5.1 x 10-9 m2 and a thickness of 1.4 x 10-8 m.

ziro4ka [17]

a) The charge on the outer surface is $1.2\cdot 10^{-12} C$

b) The number of ions is $7.5\cdot 10^6$

Explanation:

a)

The membrane behaves as a parallel plate capacitor, whose capacity is given by the equation

$C=\frac{k\epsilon_0 A}{d}$

where

k = 4.3 is the dielectric constant

$\epsilon_0 =8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

$A=5.1\cdot 10^{-9} m^2$ is the surface area

$d=1.4\cdot 10^{-8} m$ is the distance between the two plates

Substituting,

$C=\frac{(4.3)(8.85\cdot 10^{-12})(5.1\cdot 10^{-9})}{1.4\cdot 10^{-8}}=1.4\cdot 10^{-11} F$

The capacity of the membrane is related to the potential difference between the two surfaces by

$C=\frac{Q}{\Delta V}$

where here we have

Q = excess charge on one surface

$\Delta V = 85.5 mV = 0.0855 V$ is the potential difference between the two surfaces

Solving for Q, we find

$Q=C\Delta V=(1.4\cdot 10^{-11})(0.0855)=1.2\cdot 10^{-12} C$

b)

We said that the net charge on the outer surface is

$Q=1.2\cdot 10^{-12} C$

The charge of one K+ ions is equal to the electron charge

$+e=1.6\cdot 10^{-19} C$

Therefore, the number of ions on the outer surface can be found by dividing the total charge by the charge of a single ion:

$N=\frac{Q}{e}=\frac{1.2\cdot 10^{-12}}{1.6\cdot 10^{-19}}=7.5\cdot 10^6$

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2 years ago

The distance from earth to mars ranges between 780,000,000 km and 380,000,000 km depending on the time of the year. The speed of

ivann1987 [24]

Answer:

The minimum time, t = 21.11 m

Explanation:

Given,

The maximum distance between the Earth and Mars, d = 780,000,000 km

The maximum distance between the Earth and Mars, d = 780,000,000 km

The speed of the light, c = 300,000 km/s

The distance and speed of the object is related using the formula,

v = d / t

t = d / v

The minimum time it takes for data transmission is when the distance between the Mars and the Earth is at minimum.

Therefore,

t = 380,000,000 km / 300,000 km/s

= 1266.67 s

= 21.11 minutes

Hence, the minimum time it takes for data transmitted by the Mars Surveyor to reach earth, t = 21.11 minutes

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2 years ago

A binary star system consists of two equal mass stars that revolve in circular orbits about their center of mass. The period of

Vladimir79 [104]

Answer:

$m = 2.23 \times 10^{-32} kg$

Explanation:

Given data:

PERIOD OF MOTION IS T = 25.5 days

orbital speeds = 220 km/s

we know that

acceleration due to centripetal force is $a = \frac{F}{m} = \frac{V^2}{r}$

Gravitational force $F= \frac{Gm m}{d^2}$

we know that

$v = \frac{2\pi R}{T}$

solving for

$R = \frac{vT}{2\pi}$

$F = \frac{Gm^2}{4(\frac{vT}{2\pi})^2}$

$F = G\times \frac{\pi m}{(vT)^2}$

$a = \frac{v^2}{\frac{vT}{2\pi}}$

$a = \frac{2\pi v}{T}$

we know that

f =ma

$G\times \frac{\pi m}{(vT)^2} = a = \frac{2\pi m v}{T}$

solving for m

$m = \frac{2Tv^3}{\pi G}$

$m = \frac{2\times 25.5 \times 86400 \times 220000^3\ m/s}{\pi \times 6.67\times 10^{-11}}$

$m = 2.23 \times 10^{-32} kg$

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2 years ago

When you calculate the SLOPE of a line segment, what does the SLOPE represent? (Choose all that apply) the Distance traveled the

dolphi86 [110]

Answer:

Please find the answer in the explanation

Explanation:

When you calculate the SLOPE of a line segment, what does the SLOPE represent? (Choose all that apply) the Distance traveled the Displacement the Velocity the Acceleration None of the above

The slope of any time graph can not give you distance or displacement except for position - time graph.

When you plot either distance or displacement against time, that is, distance time graph or displacement time graph, you can get speed or velocity as the slope of the line segment.

You can only acceleration as a slope in a line of best fit if velocity is plotted against time. That is, in a velocity time graph.

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2 years ago

A 45.0-kg sample of ice is at 0.00°C. How much heat is needed to melt it? For water, Lf=334 kJ/kg and Lv=2257 kJ/kg

Aleonysh [2.5K]

Heat required to change the phase of ice is given by

Q = m* L

here

m = mass of ice

L = latent heat of fusion

now we have

m = 45 kg

L = 334 KJ/kg

now by using above formula

$Q = 45 * 334 * 10^3$

$Q = 1.5 * 10^7 J$

In KJ we can convert this as

$Q = 1.5 * 10^4 kJ$

so the correct answer is D option

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2 years ago

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